Symmetry of function defined by integral

Question

Define a function $f(\alpha, \beta)$, $\alpha \in (-1,1)$, $\beta \in (-1,1)$ as

$$ f(\alpha, \beta) = \int_0^{\infty} dx \: \frac{x^{\alpha}}{1+2 x \cos{(\pi \beta)} + x^2}$$

One can use, for example, the Residue Theorem to show that

$$ f(\alpha, \beta) = \frac{\pi \sin{\left (\pi \alpha \beta\right )}}{ \sin{\left (\pi \alpha\right )} \, \sin{\left (\pi \beta\right )}} $$

Clearly, from this latter expression, $f(\alpha, \beta) = f(\beta, \alpha)$. My question is, can one see this symmetry directly from the integral expression?

Answer 1

Not an answer, but a reply to Maesumi's comment --

Invariance under $\alpha \to -\alpha$ is not so difficult to see: let $x=y^{-1}$, then \begin{align*}\int_0^{\infty} \frac{x^{\alpha}}{1+2x\cos(\pi \beta) + x^2} dx &= \int_{\infty}^0 \frac{y^{-\alpha} }{1+2y^{-1}\cos(\pi\beta)+y^{-2}}(-y^{-2})dy\\ &=\int_0^{\infty} \frac{y^{-\alpha}}{1+2y\cos(\pi \beta) + y^2} dy. \end{align*}

No idea about $\alpha \leftrightarrow \beta$ though.

Answer 2

Very interesting question! But, alas, not an answer. Only few representations for the integral obtained. One of them evaluated to the form claimed in the question.

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First, transform the integral into a form, symmetric under $\alpha \mapsto -\alpha$: $$ \int_0^\infty \frac{x^\alpha}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x = \int_0^1 \frac{x^\alpha}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x + \int_1^\infty \frac{x^\alpha}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x $$ Make a change of variables $x \to x^{-1}$ in the last integral to obtain: $$ f(\alpha,\beta) = \int_0^1 \frac{x^\alpha + x^{-\alpha}}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x \tag{1} $$ Now, making a change of variables $x = \exp(-t)$ we have: $$ f(\alpha,\beta) = \int_0^\infty \frac{\cosh(\alpha t)}{\cosh(t) + \cos(\beta \pi)} \mathrm{d} t \tag{2} $$ Using $$ \int_0^\infty \exp\left(-u \left( \cosh t + \cos \pi \beta \right) \right) \mathrm{d}u = \frac{1}{\cosh(t) + \cos(\beta \pi)} $$ and the integral representation of the modified Bessel function of the second kind: $$ \int_0^\infty \cosh(\alpha t) \exp\left( - u \cosh t \right) \mathrm{d}t = K_\alpha(u) $$ we arrive at a compact representation: $$ f(\alpha,\beta) = \int_0^\infty K_\alpha(u) \mathrm{e}^{-u \cos\left(\pi \beta\right)} \mathrm{d} u \tag{3} $$ expanding the exponential into series and using $\int_0^\infty u^n K_\alpha(u) \mathrm{d} u = 2^{n-1} \Gamma\left(\frac{n}{2} + \frac{1+\alpha}{2} \right)\Gamma\left(\frac{n}{2} + \frac{1-\alpha}{2} \right)$ we get: $$ f(\alpha,\beta) = \sum_{n=0}^\infty \frac{2^{n-1}}{n!} \left(-\cos \pi \beta\right)^{n} \Gamma\left(\frac{n}{2} + \frac{1+\alpha}{2} \right)\Gamma\left(\frac{n}{2} + \frac{1-\alpha}{2} \right) \tag{4} $$ summing over even and over odd integers: $$ f(\alpha, \beta) = \frac{\pi}{2} \frac{ \cos\left( \alpha \arcsin \cos(\pi \beta) \right) }{ | \sin(\pi \beta) | \cos \left( \frac{\pi \alpha}{2} \right)} - \frac{\pi}{2} \frac{ \sin\left( \alpha \arcsin \cos(\pi \beta) \right) }{ | \sin(\pi \beta) | \sin \left( \frac{\pi \alpha}{2} \right)} = \pi \frac{\sin \left( \alpha \left( \frac{\pi}{2} - \arcsin \cos(\pi \beta) \right) \right)}{ | \sin \pi \beta | \sin(\pi \alpha)} $$ Now $\frac{\pi}{2} - \arcsin \cos(\pi \beta) = \arccos \cos(\pi \beta) = \pi | \beta |$ for $-1<\beta<1$. Thus, restoring parity, we recover the OP's expression: $$ f(\alpha, \beta) = \pi \frac{ \sin(\pi \alpha \beta)}{\sin(\pi \alpha) \sin(\pi \beta)} = \frac{\operatorname{sinc}(\pi \alpha \beta)}{\operatorname{sinc}(\pi \alpha) \operatorname{sinc}(\pi \beta)} \tag{5} $$